vector dot product help.

RadiationX

i have three vectors: a=4,b=3,c=5 that form a right triangle.
vector a is in the positive x direction, vector b is in the positive y direction starting at the tip of a. vector c is the hypotenuse of the triangle with tip at the origin. (see attaced picture .doc file)


the questions are: what is a dot b, a dot c, and b dot c.


i have the solutions in my manual but i dont understand them.


the manual says that from the figure it is clear that a + b + c = 0, where a is perpindicular to b:


a dot b = 0, since the angle between them is 90 degrees:


a dot c = a dot (-a-b)=-|a|^2=-16

and similarly b dot c = -9

i have no idea whay this is true. any help would be appreciated especially a general explanition of what the dot product is

dextercioby

Is that a joke...?You posted only 2...

Daniel.

cronxeh

A dot product between vector a and vector b is this:

a dot b = |a|*|b|*cos(theta)
a dot b = (AxBx)i + (AyBy)j + (AzBz)k

Those are two definitions and they are equal. In your case, you are given the lenght of vectors (a=4,b=3,c=5). This is the called the magnitude of a vector. the |a| = 4, |b| = 3, |c| = 5. How would you find the angle theta between the two vectors?

There are two laws you can use. Law of sines and law of cosines. Or a pythogoras' theorem if the vectors form a right angle. In your case the triangle is right, because 4^2 + 3^2 = 5^2. So you can use a good old SOH CAH TOA rule (Sin = Opposite/Hypothenus, Cos = Adjacent/Hypothenus, Tan = Opposite/Adjecent).

Try to visualize the triangle first. Obviously c is a hypothenus with lenght 5.

This is given: |a| = 4, |b| = 3, |c| = 5
And you want to find:
1] a dot b = |a|*|b|*cos(theta)= 4*3*cos(90) = 0
2] a dot c = |a|*|c|*cos(theta) = 4*5*4/5 = 16
3] b dot c = |b|*|c|*cos(theta) = 3*5*3/5 = 9

RadiationX

ok i made my mistake with soh cah toa , but the last two answers are -16 and -9. why the negative sign? thx

Doc Al

Because the angle between the vectors ([itex]\theta[/itex]) is 90 < [itex]\theta[/itex] < 180 degrees, a region in which [itex]cos \theta[/itex] is negative.

RadiationX

i'm not saying that your're wrong but how do you know that cos is in the second quad? from the picture this is not obvious.

Doc Al

To find the angle between two vectors, redraw them so that their tails start at the same point. Direction matters!

RadiationX

that's it!!! thank you Doc AI